6(d). Let E L and let R.
If = 0 then E = 0E cfw_0, so m (E) m (cfw_0) = 0, so m (E) =
0, so E L by 2(c).
If = 0, then its not hard to show (by very similar arguments to the
proof of the hints in question 5) that for A R we have
A (E) = (1 A) E) and (E)c =
MW Wu SW4 (8an
(9
LOW W943 SS Nmq 21> SUT CWLQ.
{7m , If S or T is the empty set, then this is trivial. So we may assume that S
/._. ' and T are both non-empty. Then
U0, [ON 8: {S$\,gh~§
30 SOT: gSHEHSL,E-L,§>,63)n.§
Sb SOT L5
0% Saw»: w; a, WLL. Lie.
MA2224 (Lebesgue integral) Tutorial/exercise sheet 1 Solutions
[due January 20, 2014]
1. If S = cfw_s1 , s2 , . . . is a countably innite set and t S, show that S cfw_t is also countably
/
innite.
Solution: We can list the elements of S cfw_t by inserting
MA2224 (Lebesgue integral) Tutorial sheet 2
[January 24, 2014]
Name: Solutions
1. Let I1 = (3, 5], I2 = (1, 4], I3 = (4, 2], I4 = (4, ), I5 = (9, 10].
For each of the following sets E, nd the standard representation of the set (as a union of
intervals in
MA2224 (Lebesgue integral) Tutorial sheet 4
[February 7, 2014]
Name: Solutions
1.
(a) If E = cfw_x0 R is a singleton subset (only one point), show that m (E) = 0.
Solution: Choose E1 = (x0 , x0 ] and En = for n 2. Then E1 , E2 , . . . I
and E En = E1 . S
MA2224 (Lebesgue integral) Tutorial sheet 3
[January 31, 2014]
Name: Solutions
1. If A is an algebra of subsets of R and : A [0, ] is a countably subadditive function
such that () = 0, show that it is nitely subadditive. [Recall: countably subadditive mea
MA2224 (Lebesgue integral) Tutorial sheet 5
[February 14, 2014]
Name: Solutions
1. Show that the Borel -algebra is the -algebra generated by the intervals [a, b) with a < b
(a and b nite).
Solution: By the theorem giving different collections of sets that
MA2224 (Lebesgue integral) Tutorial sheet 9
[March 28, 2014]
Name: Solutions
1. Let fn : R R be given by
2n
fn (x) =
j=1
j1
[(j1)/2n ,j/2n ) (x)
2n
(a) Show that fn = fn [0,1) for each n.
Solution: If fn (x) = 0 then x [(j 1)/2n , j/2n ) for some j = 1, 2
MA2224 (Lebesgue integral) Tutorial sheet 10
[March 28, 2012]
Name: Solutions
1. Let fn = [n,2n] .
(a) Find f (x) = limn fn (x),
lim
fn d
n
R
and
f d.
R
Solution:
f (x) = lim fn (x) = 0
n
because for any x, n > x once n is large and then fn (x) = 0.
fn d
MA2224 (Lebesgue integral) Tutorial sheet 8
[March 14, 2014]
Name: Solutions
1.
(a) Let f : R R be given by f (x) = [0,2] (x) cos x. What is f + ?
Solution: By denition f + (x) = f (x) when f (x) 0 and 0 otherwise. Since
[0,2] (x) = 0 for x [0, 2] we need
MA2224 (Lebesgue integral) Tutorial sheet 7
[March 7, 2014]
Name: Solutions
1. Verify directly that constant functions f : R (, ) are Lebesgue measurable.
Solution: Let c (, ) be the constant value so that f : R (, ) is given by
f (x) = c for each x R.
Fo
MA2224 (Lebesgue integral) Tutorial sheet 6
[February 21, 2014]
Name: Solutions
1. Let F R with F L . By we mean Lebesgue measure on L (that is the restriction
of outer measure m to L ). Dene F : L [0, ] by
F (E) = (F E)
(for E L ).
Show that F is a measu
MA2224 (Lebesgue integral) Tutorial sheet 10
[March 28, 2012]
Name: Solutions
1. Let
f=
n=1
(1)n+1
[n1,n)
n
n
n
(a) Compute
f (x) dx (Riemann) and [n,n] f d [Hint: for the Riemann integral,
divde [n, n] into sections [i, i + 1]
Solution: As a Riemann inte
MA2224 (Lebesgue integral) Tutorial/exercise sheet 1 Solutions
[due January 20, 2014]
1. If S = cfw_s1 , s2 , . . . is a countably innite set and t S, show that S cfw_t is also countably
/
innite.
Solution: We can list the elements of S cfw_t by inserting
MA2224 (Lebesgue integral) Tutorial sheet 4
[February 7, 2014]
Name: Solutions
1.
(a) If E = cfw_x0 R is a singleton subset (only one point), show that m (E) = 0.
Solution: Choose E1 = (x0 , x0 ] and En = for n 2. Then E1 , E2 , . . . I
and E En = E1 . S
MA2224 (Lebesgue integral) Tutorial sheet 3
[January 31, 2014]
Name: Solutions
1. If A is an algebra of subsets of R and : A [0, ] is a countably subadditive function
such that () = 0, show that it is nitely subadditive. [Recall: countably subadditive mea
MA2224 (Lebesgue integral) Tutorial sheet 10
[March 28, 2012]
Name: Solutions
1. Let fn = [n,2n] .
(a) Find f (x) = limn fn (x),
lim
fn d
n
R
and
f d.
R
Solution:
f (x) = lim fn (x) = 0
n
because for any x, n > x once n is large and then fn (x) = 0.
fn d
MA2224 (Lebesgue integral) Tutorial sheet 5
[February 14, 2014]
Name: Solutions
1. Show that the Borel -algebra is the -algebra generated by the intervals [a, b) with a < b
(a and b nite).
Solution: By the theorem giving different collections of sets that
MA2224 (Lebesgue integral) Tutorial sheet 2
[January 24, 2014]
Name: Solutions
1. Let I1 = (3, 5], I2 = (1, 4], I3 = (4, 2], I4 = (4, ), I5 = (9, 10].
For each of the following sets E, nd the standard representation of the set (as a union of
intervals in
MA2224 (Lebesgue integral) Tutorial sheet 8
[March 14, 2014]
Name: Solutions
1.
(a) Let f : R R be given by f (x) = [0,2] (x) cos x. What is f + ?
Solution: By denition f + (x) = f (x) when f (x) 0 and 0 otherwise. Since
[0,2] (x) = 0 for x [0, 2] we need
MA2224 (Lebesgue integral) Tutorial sheet 6
[February 21, 2014]
Name: Solutions
1. Let F R with F L . By we mean Lebesgue measure on L (that is the restriction
of outer measure m to L ). Dene F : L [0, ] by
F (E) = (F E)
(for E L ).
Show that F is a measu
MA2224 (Lebesgue integral) Tutorial sheet 10
[March 28, 2012]
Name: Solutions
1. Let
f=
n=1
(1)n+1
[n1,n)
n
n
n
(a) Compute
f (x) dx (Riemann) and [n,n] f d [Hint: for the Riemann integral,
divde [n, n] into sections [i, i + 1]
Solution: As a Riemann inte